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This is probably a very dumb question but after trying to review some calculus after years not using it, I am confused by variables in the equation for a tangent line. So I watched the very first lecture on calculus by MIT ( https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006) and at around $0:31$, I am very confused about the $X$ and $X_0$ he uses.

Basically, there is a function $y = \frac{1}{x}$ and we are trying to find the area under all triangles formed by the tangent line of this function. Differentiation was used in order to find the function $y=\frac{1}{x^2}$ which shows the slope of the original function for any $X$. I perfectly understand the logic of these steps etc. but I am extremely buffled by separating the $X$ and $X_0$ -> the lecturer even pointed out that it often confuses people but didn´t elaborate more on it.

Basically, the problem for me starts right when trying to find the equation for the tangent line, given the derivative we found. The equation for a line is $Y - y_0 = m(X-x_0)$ For some reason, I am tempted to just write: $Y - y_0 = (\frac{1}{x^2})*(X-X)$ which obviously is a nonsense since we´d obtain a zero in the parenthasis. But I just cannot justify in my mind the fact that we use x0 instead of $X$ to model the slope $m$. I´d say that the slope will be a function of $X$ and therefore, I´d never ever think of using something like the $x_0$. Can please someone help to clarify this for me? thank you!

UNAN
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kuchejdatomas
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    Hi, welcome. I couldn't find the video that you're referring to when I clicked the link you gave. Can you give a more direct one? – Matthew Leingang Jul 07 '21 at 16:30
  • Hi, sure, here it is: https://www.youtube.com/watch?v=jbIQW0gkgxo&ab_channel=MITOpenCourseWare

    The problem starts at 27:18

     – kuchejdatomas Jul 07 '21 at 16:51
  • Oh I see it now! In the original post you wrote 0:31 so I thought you mean 31 seconds in. Now I understand you mean 31 minutes in. – Matthew Leingang Jul 07 '21 at 16:55

2 Answers2

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Well, the underlying concept is simple. We take a certain point $(x_0, y_0)$ lying on the given curve. At this point, slope of tangent is obviously given by: $$m=\left(\frac {dy}{dx}\right)_{x=x_0}$$ Hence, for your given curve, at $x=x_0$, $$m=-\frac {1}{{x_0}^2}$$ Thus, keeping in mind that $(x_0, y_0)$ are not variables, rather parameters, we can write equation of tangent: $$\frac {y-y_0}{x-x_0}=m=-\frac {1}{{x_0}^2}$$ Here $x,y$ represent the variables in which we represemt the line.The slope, $m$, and one point on the line, that is, the point of tangency $(x_0,y_0)$, parameters for us.

Ritam_Dasgupta
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  • What's the difference between a variable and a parameter? – Matthew Leingang Jul 07 '21 at 16:53
  • Treating x0,y0 as parameters makes so much more sense. Basically, all I am confused about is when one should use X as a general variable and when x0 as a parameter in this case. The temptation is to use the general variable X everywhere instead of the x0... I hope you understand – kuchejdatomas Jul 07 '21 at 16:54
  • A parameter is simply an unknown constant, whereas a variable has a non-constant value. – Ritam_Dasgupta Jul 07 '21 at 17:00
  • If you are indeed so confused using $x_0$ and $y_0$, replace them with $h$ and $k$. That is standard terminology used when finding locii. – Ritam_Dasgupta Jul 07 '21 at 17:02
  • I disagree with this answer to an extent. When we write $y=x^2$, it's not that $x$ and $y$ are literally varying either. This notation simply describes the set of points ${(x,y):y=x^2,x\in\Bbb{R}}$. The idea of a "variable quantity" was used by the creators of calculus, but the more modern notion of a variable is of a fixed quantity that represents any given member of a set. – Joe Jul 07 '21 at 18:35
  • I simply meant that for the equation of the line, $x_0$ and $y_0$ are to be treated as unknown constants, whereas $x,y$ are the variables. – Ritam_Dasgupta Jul 07 '21 at 19:19
  • Thanks a lot. Do you by any chance know about any resources (like khan academy) where I could learn more & practice this concept of variable vs. constant? It seems to be such an important topic but at the same time it can be quite confusing - especially if one has experience from working with variables in various programming languages, where variables have a bit different chracteristics.

    The reason I am still tempted to use m = 1/X^2 instead of m = 1/X_0 is that m actually is a function of X. In the end, I still end up plugging numbers from the X axis into the X_0 values anyway...

     – kuchejdatomas Jul 07 '21 at 22:12
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I'm getting nostalgia for lecturing with chalk and erasers like those...

You write, “we are trying to find the area under all triangles formed by the tangent line of this function.” But there's not just a single tangent line to the curve, is there? There are infinitely many—any point on the curve is going to produce a tangent line to the curve at that point. The professor says:

We're going to calculate the areas of triangles, and you could ask yourself, “In terms of what?”

He proceeds to mark a generic point on the curve with coordinates $(x_0,y_0)$. Now the goal is much more refined: find the area of of the triangle enclosed by the axes and line tangent to the curve $y=1/x$ at $(x_0,y_0)$, as an expression involving (“in terms of”) $x_0$ and $y_0$.

Why not label that point $(x,y)$ instead of $(x_0,y_0)$? Because the professor wants to use $x$ and $y$ as variables for the equation of the curve, and the equation for the tangent line(s). As you discovered, if you try to do both, your algebraic expressions collapse to nothing as a result of this overloading.

After the example is finished the professor points out what he sees as a big obstacle to learning calculus: understanding what the variables are, and when. As Ritam Dasgupta notes in his answer, $x_0$ isn't changing in the course of this problem—it's unknown, but it's fixed. In this particular problem the area of the triangle is a constant, no matter what tangent line is used, but you can imagine curves where this area depends on $x_0$. What if you wanted to the largest triangle? Well, now you do treat $x_0$ like a variable and differentiate the area function in terms of it.

Don't assume that if this example is confusing, you're not meant to learn calculus. I'm not sure this example is the best choice for the first day of Calculus I. MIT led the way on open courseware back in 2006, but by now there are many, many calculus lectures online. Use that to your advantage and find one that's to your liking.

Matthew Leingang
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  • Added to this problem, the traditional notation for derivatives, where we write $$ \frac{dy}{dx}=-\frac{1}{x^2} , , $$ means that at each point $(x,y)$ on the curve $y=1/x$, the change to the $y$-coordinate is $-1/x^2$ times the change in the $x$-coordinate. Arguably $x$ and $y$ are being used in two different ways here, though usually this doesn't cause too many problems in practice. – Joe Jul 07 '21 at 18:43
  • Thank you very much for the explanation! Well, I think my confusion stems from being used to treating variables differently in programming (python). When I see X, I see a vector of all the real numbers on the horzontal axis - an object. This is why I am tempted to plug it in the slope fraction m instead of the X_0. The variables X,Y as used here work quite differently, which confuses me. – kuchejdatomas Jul 07 '21 at 18:47
  • @kuchejdatomas Oh, wow, I can see how that perspective would be confusing. I've never thought about variables that way. Maybe I can ponder more on that and update the answer. – Matthew Leingang Jul 07 '21 at 18:57